But this doesn't answer the second part of the question: Are there any solutions to the differential equation that are missing from the set of solutions you found?

Are there any solutions to the DE that are missing from the set (1)?

Your "solution" is not a solution at all; a solution will not have a [itex] \pm [/itex] in it. You need to pick either [tex] y_1(t) = \sqrt{t \ sin (2t)+ \frac{1}{2} \ cos (2t) + k_1} \text{ or } y_2(t) = -\sqrt{t \ sin (2t)+ \frac{1}{2} \ cos (2t) + k_2}.[/tex] (If you don't believe me, try plugging in you written solution into a program such as Maple and see whether or not the soflware can handle it.) That being said, the question becomes: are there any solutions different from [itex] y_1 \text{ or } y_2[/itex]? You need to use some theorems, for example, related to the IVP, where in addition to the DE you also specify y(0) for example.

Your "solution" is not a solution at all; a solution will not have a [itex] \pm [/itex] in it. You need to pick either [tex] y_1(t) = \sqrt{t \ sin (2t)+ \frac{1}{2} \ cos (2t) + k_1} \text{ or } y_2(t) = -\sqrt{t \ sin (2t)+ \frac{1}{2} \ cos (2t) + k_2}.[/tex] (If you don't believe me, try plugging in you written solution into a program such as Maple and see whether or not the soflware can handle it.) That being said, the question becomes: are there any solutions different from [itex] y_1 \text{ or } y_2[/itex]? You need to use some theorems, for example, related to the IVP, where in addition to the DE you also specify y(0) for example.

So we must when we are solving a DE and have a situation like y2=..., we must always take the positive value?

1. If using the "±" symbol is wrong, then what is the correct notation for representing the solution to this DE?

2. So, there are absolutely no missing solutions, and my answer is the general solution and we can solve all initial-value problems with solutions of this form? (Personally I can't think of any counter examples, but I'm not sure).

So we must when we are solving a DE and have a situation like y2=..., we must always take the positive value?

Why would you ask me that, when I very clearly wrote down TWO solutions?

Anyway, if you look at the IVP and have y(0) > 0, then you must choose the solution y1 (with the + sign); if y(0) < 0 you must choose y2 (with the - sign). Then the issue is: are there any other solutions to your IVP? You can use theorems about that. Google "uniqueness theorems for differential equations".

1. If using the "±" symbol is wrong, then what is the correct notation for representing the solution to this DE?

i didn't say it was wrong!

if your professor doesn't like it, don't do it

if your professor doesn't mind it, do do it

it's a convenient shorthand …

when we write "the general solution is x = Acoskt + C", we don't feel the need to add "for any constant C"

when we write "the general solution is x = -b ±√(b2 - 4c), we don't feel the need to add "for either value of ±"

2. So, there are absolutely no missing solutions, and my answer is the general solution and we can solve all initial-value problems with solutions of this form? (Personally I can't think of any counter examples, but I'm not sure).

when we write "the general solution is x = Acoskt + C", we don't feel the need to add "for any constant C"

when we write "the general solution is x = -b ±√(b2 - 4c), we don't feel the need to add "for either value of ±"

looks ok

The point I was trying to get across to him was that y = ± f(t) is _not_ a solution; it is TWO DIFFERENT solutions, but written in shorthand. As I suggested to him, try submitting his "solution" y = ± f(t) to Maple, Mathematica or Matlab to test it. The computer would choke. I know, I know that we sometimes say "the root is a ± b" and similar language, but *if one does not fully understand this one should avoid it*! It is really a shorthand form of the statement that there are two roots, a + b and a - b. It is perfectly acceptable to say "the roots are a ± b"; this emphasizes roots (plural) and is absolutely correct in all respects.

When y=0, the partial derivative fails to exist, this means that the uniqueness theorem doesn't tell us anything about solutions of an IVP that have the form y(t0)=0. For example at the point y(0)=0, neither f(t, y) or ∂f/∂y exist. But how can we use this to prove that y(0)=0 is another solution to the IVP?

I'm confused because the satisfaction of the theorem simply guarantees a uniques solution, but failure of the theorem doesn't prove multiple solutions (inconclusive). So how can we use this theorem?

So, if y(0) is not zero you have a unique solution; in fact, you can see from the DE itself that y(t) is strictly increasing if y(0) > 0 and is strictly decreasing if y(0) < 0, so y(t) never reaches zero. The case y(0) = 0 is pathological, and I don't think you can say the DE even means anything right at the point t = 0.

So f(t, y) is not defined at y=0, but it is continious everywhere else and ∂f/∂y is continious everywhere, and by the Uniqueness Theorem the solution to every initial-value problem is unique. Is this a sufficient explanation that there are no missing solutions?

The Uniqueness Theorem says if f(t,y) and ∂f/∂y are BOTH continuous functions of t and y for all t and y, then there is a unique solution to the DE for any particular choice of the initial condition.

To show that there are no missing solutions I must show thate there are NO values of t and y at which either f(t,y) and ∂f/∂y is discontinuous. BUT at the point y=0 the function f(t,y) is not defined/has a discontinuity. So one of the hypotheses of the theorem is violated. Doesn't this mean there is a missing solution?

The Uniqueness Theorem says if f(t,y) and ∂f/∂y are BOTH continuous functions of t and y for all t and y, then there is a unique solution to the DE for any particular choice of the initial condition.

To show that there are no missing solutions I must show thate there are NO values of t and y at which either f(t,y) and ∂f/∂y is discontinuous. BUT at the point y=0 the function f(t,y) is not defined/has a discontinuity. So one of the hypotheses of the theorem is violated. Doesn't this mean there is a missing solution?

No. The conditions you cite are *sufficient* for uniqueness, not necessary. In other words, even if the conditions are violated that does not mean that uniqueness fails; in some examples it will, and in other examples it won't.

You KNOW that the DE is meaningless at y = 0, so you could say that it must hold if y ≠ 0. That does not mean that y(t)=0 is impossible; it just means that the DE fails at such a point. For example, the solution [itex]y(t) = \sqrt{t \sin(2t) + \cos(t)/2 - 1/2} [/itex]
gives [itex] y(t) \rightarrow 0 \text{ as } t \rightarrow 0,[/itex] and [itex] y(0) = 0,[/itex] so y(t) is continuous at t = 0. The DE holds for all t > 0 but not at t = 0 itself.